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hjlf
3 years ago
7

Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other in witch states coul

d mercury naturally exits?
Physics
2 answers:
Thepotemich [5.8K]3 years ago
7 0

Answer:

In nature, there are 3 common states.

Solid, gas and liquid.

The solids have the atoms really close to each other and in fixed positions, so they have a defined shape, the gases have the atoms far away from each other and are really free, so they only can have a "defined shape" when are inside a closed container. And the liquids have the atoms close to each other but are free, those can take the shape of open containers, for example, water in a glass.

Now, the natural state of mercury is the liquid state at room temperature(you can see this in a mercury thermometer, for example.

And of course, depending on the temperature and pressure this state can change, for example, if the temperature is low enough, the mercury will take a solid-state.

Stels [109]3 years ago
5 0

Mercury's natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) could mercury naturally exist?

Liquid is the answer

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Answer:

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<em> </em><em>?</em><em>:</em><em>:</em><em>:</em><em> </em><em>=</em><em>1</em>

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6 0
3 years ago
Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o
sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

U = \frac{kq_{1}q_{2}  }{r}

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

U_{1}  = \frac{ke^{2}   }{r_{1} }

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

U_{2}  = \frac{ke^{2}   }{r_{2} }=\frac{ke^{2}   }{4.10r_{1} }

Applying law of conservation of energy :

ΔU  = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁  - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

\frac{1}{2}mv^{2}=\frac{ke^{2}   }{r_{1} }- \frac{ke^{2}   }{4.10r_{1} }

Here m and v are the mass and final speed of electron respectively.

v^{2}=\frac{2}{m} \frac{ke^{2}   }{r_{1} }(1- \frac{1  }{4.10 })

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}   }{4.32\times10^{-10} }(1- \frac{1  }{4.10 })

v^{2}=8.86\times10^{11}

v = 9.41 x 10⁵ m/s

5 0
2 years ago
How do i calculate this?
Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

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41.5 is the answer that i got. hope this helps!

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